Work, Energy and Power Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1(a) Positive: the man pulls the bucket upward and it moves upward — force and displacement are in the same direction.
2. 2(b) Negative: gravity acts downward while the bucket moves upward — force and displacement are opposite.
3. 3(c) Negative: friction opposes the motion of the body sliding down.
4. 4(d) Positive: to keep a uniform velocity on a rough plane an applied force must act forward — it acts in the direction of motion. (Friction, of course, does negative work.)
5. 5(e) Negative: the resistive force of air opposes the direction of motion of the pendulum.

Step-by-step solution

1. 1m = 2 kg, F = 7 N, μ = 0.1, u = 0, t = 10 s.
2. 2Friction: f = μmg = 0.1 × 2 × 9.8 = 1.96 N (opposing motion).
3. 3Net force F_net = 7 − 1.96 = 5.04 N; acceleration a = 5.04/2 = 2.52 m s⁻².
4. 4Distance travelled: s = ½at² = ½ × 2.52 × (10)² = 126 m.
5. 5(a) W_applied = F·s = 7 × 126 = 882 J.
6. 6(b) W_friction = −f·s = −1.96 × 126 = −247 J.
7. 7(c) W_net = F_net·s = 5.04 × 126 = 635 J.
8. 8(d) v = at = 2.52 × 10 = 25.2 m s⁻¹; ΔKE = ½mv² − 0 = ½ × 2 × (25.2)² = 635 J.
9. 9Interpretation: the net work (635 J) equals the change in kinetic energy — the work-energy theorem. The applied work minus the frictional work gives the same 635 J.

Step-by-step solution

1. 1Since E = V + K and kinetic energy cannot be negative, the particle cannot exist wherever V exceeds E.
2. 2(a) V rises steeply and crosses E at x = a: the particle cannot be found for x > a. Minimum total energy = 0.
3. 3(b) V = V₀ is greater than E everywhere: the particle cannot be found in any region. Minimum total energy = V₀.
4. 4(c) The well falls to −V₁ between x = a and x = b; outside this region V > E. Forbidden: x < a and x > b. Minimum total energy = −V₁.
5. 5(d) The double well touches −V₁ at two minima. V exceeds E in the central hump between the two wells and outside the wells. Forbidden: the middle region and x < −b/2, x > b/2. Minimum total energy = −V₁.
6. 6Physical contexts: (a) a particle near a repulsive wall, (b) motion confined by a high barrier, (c) a simple potential well (an attached mass on a spring track), (d) a double well (a diatomic molecule's asymmetric stretch about a central maximum).

Step-by-step solution

1. 1At the turning points the particle momentarily stops: K = 0, so E = V.
2. 2E = ½kx² with k = 0.5 N m⁻¹ and E = 1 J.
3. 31 = ½ × 0.5 × x² = 0.25 x².
4. 4x² = 4, so x = ±2 m.
5. 5Beyond x = ±2 m, V would exceed E, which is impossible; hence the particle must turn back at ±2 m.

Step-by-step solution

1. 1(a) From the rocket. The burning reduces the rocket's own mass, and its total energy (mgh + ½mv²) drops by the amount burned.
2. 2(b) Gravitation is a conservative force: work over any closed path (a complete orbit) is zero.
3. 3(c) As the orbit shrinks the potential energy decreases; the total energy stays essentially constant, so the lost P.E. appears as kinetic energy — the speed rises. (The thin drag only slowly drains the total energy; its dominant effect near the earth is a steeper dive, not a slowing.)
4. 4(d) In case (i) the man's supporting force is vertical while the displacement is horizontal: W = mgs cos 90° = 0.
5. 5In case (ii) the rope is pulled in its own direction: W = mgs = 15 × 9.8 × 2 = 294 J.
6. 6So the work done is greater in the second case — pulling the rope.

Step-by-step solution

1. 1(a) Decreases: positive work by a conservative force moves the body along the force, reducing separation from the centre of force.
2. 2(b) Kinetic: friction acts against motion and the work done drains kinetic energy.
3. 3(c) External force: internal forces always cancel in pairs and cannot change total momentum.
4. 4(d) Total linear momentum: momentum is conserved in every collision, elastic or inelastic (kinetic energy is not).

Step-by-step solution

1. 1(a) False: in an elastic collision the total momentum and total energy of the two bodies are conserved, not necessarily the energy or momentum of each individual body.
2. 2(b) False: external forces can do work and change the energy of a system; energy is not conserved when unbalanced external forces act.
3. 3(c) False: closed-loop work is zero only for a conservative force (e.g. gravity, spring force), not for friction or other dissipative forces.
4. 4(d) True: inelastic collisions lose energy to heat, sound, deformation, so the final kinetic energy is always less than the initial.

Step-by-step solution

1. 1(a) No: during contact the balls deform and some kinetic energy is briefly stored as elastic potential energy; only the total (KE + PE) is momentarily conserved.
2. 2(b) Yes: no external horizontal force acts during the collision, so linear momentum is conserved throughout.
3. 3(c) For an inelastic collision: (a) becomes "total kinetic energy is not conserved" — some is permanently lost; (b) stays yes — momentum is always conserved.
4. 4(d) Elastic: a force that depends only on separation is conservative, so the collision stores and returns energy — the result is an elastic collision.

Step-by-step solution

1. 1With constant acceleration a, the force F = ma is constant.
2. 2The velocity grows as v = at, so v ∝ t.
3. 3Power P = F·v, with F constant and v ∝ t: P ∝ t.
4. 4Hence option (ii).

Step-by-step solution

1. 1P = F·v = mav = mv(dv/dt) = constant.
2. 2So v dv = (P/m) dt; integrating: v²/2 = (P/m)t, i.e. v = √(2Pt/m).
3. 3v = dx/dt ∝ t¹ᐟ².
4. 4Integrating again: x ∝ t³ᐟ².
5. 5Hence option (iii).

Step-by-step solution

1. 1Displacement s = 4 m along z: s = 4k̂ m.
2. 2W = F·s = (−î + 2ĵ + 3k̂)·(4k̂).
3. 3The î and ĵ components are perpendicular to s, so they contribute nothing: W = 3 × 4 = 12 J.

Step-by-step solution

1. 1E_Ke = 10 keV = 1.60 × 10⁻¹⁵ J; E_Kp = 100 keV = 1.60 × 10⁻¹⁴ J.
2. 2v_e = √(2E_Ke/m_e) = √(2 × 1.60 × 10⁻¹⁵ / 9.11 × 10⁻³¹) ≈ 5.93 × 10⁷ m s⁻¹.
3. 3v_p = √(2E_Kp/m_p) = √(2 × 1.60 × 10⁻¹⁴ / 1.67 × 10⁻²⁷) ≈ 4.38 × 10⁶ m s⁻¹.
4. 4The electron is faster.
5. 5v_e/v_p = (5.93 × 10⁷)/(4.38 × 10⁶) = 13.54.

Step-by-step solution

1. 1r = 2 × 10⁻³ m; volume V = (4/3)πr³; density of water ρ = 10³ kg m⁻³.
2. 2m = ρV = (4/3) × 3.14 × (2 × 10⁻³)³ × 10³ kg.
3. 3First half (h = 250 m): W₁ = mgh = 0.082 J.
4. 4Second half: gravitational force is the same and the fall is again 250 m, so W₂ = 0.082 J.
5. 5Without resistance, total energy left over at the ground would be mg × 500 = 0.164 J.
6. 6With resistance the drop lands at 10 m s⁻¹, so final kinetic energy = ½mv² = 1.675 × 10⁻³ J.
7. 7Work by the resistive force = E_ground − E_top = (−0.164) + 0.001675 ≈ −0.162 J.

Step-by-step solution

1. 1Momentum is conserved in every collision — here the (very massive) wall recoils a negligible amount while the molecule's momentum change is absorbed by wall + earth.
2. 2The molecule rebounds with the same speed 200 m s⁻¹.
3. 3Speed is unchanged, so kinetic energy is conserved: the collision is elastic.

Step-by-step solution

1. 1Mass of water m = ρV = 10³ × 30 = 3 × 10⁴ kg; t = 900 s; h = 40 m.
2. 2Useful (output) power P₀ = mgh/t = (3 × 10⁴ × 9.8 × 40)/900 = 13.07 × 10³ W.
3. 3Efficiency η = P₀/Pᵢ = 30%.
4. 4Pᵢ = P₀/0.30 = (13.07 × 10³)/0.30 ≈ 4.36 × 10⁴ W = 43.6 kW.

Step-by-step solution

1. 1Check momentum and kinetic energy for each proposed result. (Mass of each ball = m.)
2. 2Before: momentum = mV; KE = ½mV².
3. 3Result (i): the striking ball stops and the two move together with V/2. Momentum mV = (2m)(V/2) ✓ but KE = ½(2m)(V/2)² = ¼mV² ≠ ½mV² ✗ — possible result.
4. 4Result (ii): the striking ball stops; the balls separate, one moves off with speed V. Momentum mV ✓ and KE = ½mV² ✓.
5. 5Result (iii): all three move together with V/3. KE = ½(3m)(V/3)² = mV²/6 ≠ ½mV² ✗.
6. 6Only result (ii) conserves both momentum and kinetic energy — it is the possible elastic outcome.

Step-by-step solution

1. 1For an elastic head-on collision of two equal masses, one at rest: the moving mass comes to rest and the stationary mass moves off with the full incoming velocity.
2. 2Bob B (initially at rest) receives the entire velocity of A.
3. 3Bob A comes to rest at the point of collision, so it rises by zero height.

Step-by-step solution

1. 1Initial energy (horizontal position): E = mgl = m × 9.8 × 1.5.
2. 25% is dissipated, so the energy at the lowermost point is 95% of E.
3. 3½mv² = 0.95 × m × 9.8 × 1.5.
4. 4v² = 2 × 0.95 × 9.8 × 1.5 = 27.93.
5. 5v ≈ 5.28 m s⁻¹.

Step-by-step solution

1. 1The system (trolley + sandbag) is on a frictionless track, so no external horizontal force acts on it.
2. 2Sand falls vertically out of the hole, carrying no horizontal momentum with it — it separates with the same forward velocity as the trolley.
3. 3By Newton's first law / conservation of momentum, the trolley's horizontal velocity is unchanged.
4. 4Hence the speed remains 27 km/h even after the bag is empty.

Step-by-step solution

1. 1At x = 0: v = 0, so u = 0.
2. 2At x = 2 m: v = a x³ᐟ² = 5 × 2³ᐟ² = 5 × 2√2 = 10√2 m s⁻¹.
3. 3By the work-energy theorem: W = ΔKE = ½m(v² − u²).
4. 4W = ½ × 0.5 × (200 − 0) = 50 J.

Step-by-step solution

1. 1(a) Volume swept in time t: A v t; mass m = ρ A v t.
2. 2(b) Kinetic energy of that air = ½ m v² = ½ ρ A v³ t.
3. 3(c) v = 36 km/h = 10 m s⁻¹; electrical energy = 25% of wind kinetic energy = (¼) × ½ ρ A v³ t = (1/8) ρ A v³ t.
4. 4Electrical power = energy/time = (1/8) ρ A v³.
5. 5P = (1/8) × 1.2 × 30 × (10)³ = 4500 W = 4.5 kW.

Step-by-step solution

1. 1(a) W = n m g h = 1000 × 10 × 9.8 × 0.5 = 49,000 J = 49 kJ.
2. 2(b) Usable energy from 1 kg of fat = 20% × 3.8 × 10⁷ = 7.6 × 10⁶ J.
3. 3Fat consumed = 49,000 / 7.6 × 10⁶ ≈ 6.45 × 10⁻³ kg.

Step-by-step solution

1. 1(a) Useful power per square metre = 20% × 200 = 40 W m⁻².
2. 2Area needed A = 8000/40 = 200 m².
3. 3(b) 200 m² is roughly a 14 m × 14 m roof — comparable to (a little larger than) the roof of a typical house.